Verified Solution • 2025–26 Session moderate

QJEEM23Ma78a0: JEE Main 2023 Mathematics - Class 11 JEEM23Ma78a0 Solution: The equation has :

The equation e4x+8e3x+13e2x−8ex+1=0,x∈R\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R} has :

Multiple Choice Options

  • A two solutions and both are negative
  • B two solutions and only one of them is negative
  • C four solutions two of which are negative
  • D no solution

Step-by-Step Verified Solution

e4x+8e3x+13e2x−8ex+1=0e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0

Let ex=t\mathrm{e}^{\mathrm{x}}=\mathrm{t}

Now, t4+8t3+13t2−8t+1=0\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0

Dividing equation by t2\mathrm{t}^{2}

amp;t2+8t+13−8t+1t2=0amp;t2+1t2+8(t−1t)+13=0amp;(t−1t)2+2+8(t−1t)+13=0\begin{aligned}& t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\& t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\\\& \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0\end{aligned}

Let t−1t=z\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}

z2+8z+15=0z^{2}+8 z+15=0

amp;(z+3)(z+5)=0\begin{aligned}& (z+3)(z+5)=0\end{aligned}

z=−3 or z=−5z=-3 \text { or } z=-5

So, t−1t=−3\mathrm{t}-\frac{1}{\mathrm{t}}=-3 or t−1t=−5\mathrm{t}-\frac{1}{\mathrm{t}}=-5

t2+3t−1=0t^{2}+3 t-1=0 or t2+5t−1=0t^{2}+5 t-1=0

t=−3±132\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2} or t=−5±292\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}

as t=ext=e^{x} so tt must be positive,

t=13−32 or 29−52t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}

So, x=ln⁡(13−32)x=\ln \left(\frac{\sqrt{13}-3}{2}\right) or x=ln⁡(29−52)x=\ln \left(\frac{\sqrt{29}-5}{2}\right)

Hence two solutions and both are negative.

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